Thank you for reporting, we will resolve it shortly
Q.
The energy of a hydrogen atom in the ground state is $-13.6\, eV$. The energy of a $He^+$ ion in the first excited state will be
AIPMTAIPMT 2010Atoms
Solution:
Energy of an hydrogen like atom like He$^+$ in an n$^{th}$ orbit is given by
$E_n =-\frac{13.6Z^2}{n^2}eV$
For hydrogen atom, $Z = 1$
$\therefore E_n =-\frac{13.6}{n^2}eV$
For ground state, $n = 1$
$\therefore E_1=-\frac{13.6}{1^2}eV=-13.6\,eV$
For $He^{+} ion , Z=2$
$E_n=-\frac{4(13.6)}{n^2}eV$
For first excited state, n = 2
$\therefore E_2=-\frac{4(13.6)}{(2)^2}eV=-13.6\,eV$
Hence, the energy in $He^+$ ion in first excited state is same that of energy of the hydrogen atom in ground state i.e.$ - 13.6 \,eV$